Question: A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its position vector is $(-t^2+10t,t^3-10t)$. What is the particle's velocity vector at $t=4$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(24,24)$ (Choice B) B $(-8,8)$ (Choice C) C $(8,-32)$ (Choice D) D $(2,38)$
Solution: Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's position vector is $(-t^2+10t,t^3-10t)$. We are asked to find the particle's velocity vector at $t=4$. In other words, we need to find $\vec{v}(4)$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(-t^2+10t),\dfrac{d}{dt}(t^3-10t)\right) \\\\ &=(-2t+10,3t^2-10) \end{aligned}$ Finding $\vec{v}(4)$ $\begin{aligned} \vec{v}({4})&=(-2({4})+10,3({4})^2-10) \\\\ &=(-8+10,48-10) \\\\ &=(2,38) \end{aligned}$ In conclusion, the particle's velocity vector at $t=4$ is $(2,38)$.